A potentiometer wire $PQ$ of $1\,m$ length is connected to a standard cell $E _{1}$. Another cell $E _{2}$ of emf $1.02\, V$ is connected with a resistance $'r'$ and switch $S$ (as shown in figure). With switch $S$ open, the null position is obtained at a distance of $49\, cm$ from $Q$. The potential gradient in the potentiometer wire is.......$V/cm$
JEE MAIN 2020, Medium
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Sol. Balancing length is measured from $P.$
So $100-49=51 cm$
$E _{2}=\phi \times 51$
Where $\phi=$ Potential gradient
$1.02=\phi \times 51$
$\phi=0.02 V / cm$
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